Solving $\sum_{i=3}^{n+1} i$

Question

trying to solve $\sum_{i=3}^{n+1} i$

First I attempt to change the lower/upper bounds:

$\sum_{i=3}^{n+1} i = \sum_{i=1}^{n-1} i$

in order to use $\sum_{i=1}^{n} i = \frac{n(n+1)}2$

so, $\sum_{i=1}^{n-1} i = \frac{(n-1)(n-1+1)}2 = \frac{n^2-n}2$

This is just practice out of a textbook that doesn't have answers - but I tried to input the summation in wolframalpha and my result is not one of the answers there. Where have I messed up?

Additionally, is modifying the lower/upper bound of a summation in order to use an equality like the one above an ok way to approach these problems?

Answer 1

You are probably making things complicate. It is nothing but $$ \biggr(\sum_{k=1}^nk\biggr)-1-2+(n+1) $$ and you have already known the formula for $\sum_{k=1}^nk$.

Answer 2

That's wrong. Take $n=2$. Then you say $3=1$.

Use $$ \sum_{i=3}^{n+1} i = \sum_{i=1}^{n+1} i-\sum_{i=1}^{2} i $$

If you want change order of summation, then set $m=i-2$. Then $m$ goes from $1$ to $n-1$ but $i=m+2$ so $$ \sum_{i=3}^{n+1} i=\sum_{m=1}^{n-1} (m+2) $$

Answer 3

When you subtract 2 from the lower and upper bounds, you have to replace $i$ with $(i+2)$ wherever it appears in the summands to compensate.

$\sum_{i=3}^{n+1} i=\sum_{i=1}^{n-1} (i+2)=\left (\sum_{i=1}^{n-1} i\right)+\left (\sum_{i=1}^{n-1} 2\right)$.

Can you take things from here?